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JEE Main & Advanced Mathematics Differentiation Question Bank Self Evaluation Test - Limits and Derivatives

  • question_answer
    If \[{{x}_{1}}=3\] and \[{{x}_{n+1}}=\sqrt{2+{{x}_{n}},}n\ge 1,\] then \[\underset{n\,\to \,\infty }{\mathop{\lim }}\,{{x}_{n}}\] is equal to

    A) \[-1\]

    B) \[2\]

    C) \[\sqrt{5}\]

    D) \[3\]

    Correct Answer: B

    Solution :

    [b] \[{{x}_{n+1}}=\sqrt{2+{{x}_{n}}}\Rightarrow \lim {{x}_{n+1}}=\sqrt{2+\lim {{x}_{n}}}\] \[\Rightarrow t=\sqrt{2+t}\]         \[[\because \lim {{x}_{n+1}}=lim{{x}_{n}}]\] \[\Rightarrow {{t}^{2}}-t-2=0\Rightarrow (t-2)(t+1)=0\Rightarrow t=2.\]


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